Problem: A curve in the plane is defined parametrically by the equations $x=2\cos(3t)$ and $y=-3\sin(2t)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\cos(2t)}{\sin(3t)}$ (Choice B) B $-6\cos(2t)$ (Choice C) C $-\csc^2(6t)$ (Choice D) D $\dfrac{\cos(2t)}{\sin(3t)}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=2\cos(3t)$ and $y=-3\sin(2t)$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}(-3\sin(2t))}{\dfrac{d}{dt}(2\cos(3t))} \\\\ &=\dfrac{-6\cos(2t)}{-6\sin(3t)} \\\\ &=\dfrac{\cos(2t)}{\sin(3t)} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\cos(2t)}{\sin(3t)}$.